At $300 K$ , two pure liquids $A$ and $B$ have vapour pressures $150 m m H g$ and $100 m m H g$ respectively. In an equimolar liquid mixture of $A$ and $B$ , the mole fraction of $B$ in the vapour mixture at this temperature is:

0.6

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0.5

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0.8

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0.4

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Solution

The correct option is D $0.4$
Let the moles of $A = x$

$∴$ The moles of $B = x [∵$ mixture is equimolar.]

Mole fraction of $A = \frac{x}{x + x} = 0.5 m o l e$

$=$ mole fraction of $B$

Thus, $ρ_{T} = ρ_{A}^{0} χ A + ρ_{B}^{0} χ B$

$ρ_{T} = 150 \times 0.5 + 100 \times 0.5 = 125$

$∴$ Mole fraction of $B$ in the vapour mixture $= \frac{ρ_{B}^{0} χ_{B}}{ρ_{T}}$

$= \frac{100 \times 0.5}{125} = 0.4$ .

Hence, option D is correct.

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At 300 K, two pure liquids A and B have vapour pressures 150 mm Hg and 100 mm Hg respectively. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is:

The correct option is D 0.4Let the moles of A=x∴ The moles of B=x[∵ mixture is equimolar.]Mole fraction of A=xx+x=0.5 mole= mole fraction of BThus, ρT=ρ0AχA+ρ0BχBρT=150×0.5+100×0.5=125∴ Mole fraction of B in the vapour mixture =ρ0BχBρT =100×0.5125=0.4.Hence, option D is correct.

At $300 K$ , two pure liquids $A$ and $B$ have vapour pressures $150 m m H g$ and $100 m m H g$ respectively. In an equimolar liquid mixture of $A$ and $B$ , the mole fraction of $B$ in the vapour mixture at this temperature is: